Advanced book on Mathematics Olympiad

370 Algebra 112.Bring the denominator to the left: (a+b)(b+c)(c+a) ( 1 a+b + 1 b+c + 1 c+a + 1 23 √ abc ) ≥(a+b+c+^3 √ abc)^2. T ...

Algebra 371 a d b c a c b d a d b c Figure 59 Foru/∈D 2 , the triangle inequality gives |u−λi|≥|u−c|−|c−λi|>R+|k|−R=|k|. Henc ...

372 Algebra This is equivalent to 0 <(a+b+c)(−a+b+c)(a−b+c)(a−b−c). It follows that−a+b+c,a−b+c,a−b−care all positive, becaus ...

Algebra 373 ∥ ∥∥ ∥∥ (m ∏ i= 1 Vi ) x− (m ∏ i= 1 Wi ) x ∥ ∥∥ ∥∥=‖VVmx−WWmx‖ =‖V(Vm−Wm)x+(V−W)Wmx‖. Now we use the triangle inequa ...

374 Algebra 120.We start with the algebraic identity x^3 (y−z)+y^3 (z−x)+z^3 (x−y)=(x+y+z)(x−y)(y−z)(z−x), wherex, y, zare compl ...

Algebra 375 and the inequality is proved. (64th W.L. Putnam Mathematical Competition, 2003) 124.The inequality from the statemen ...

376 Algebra 126.The solution is based on the Lagrange identity, which in our case states that ifMis a point in space andGis the ...

Algebra 377 and each of these is just the AM–GM inequality. (Greek Team Selection Test for the Junior Balkan Mathematical Olympi ...

378 Algebra x 1 −^1 n 1 +x 1 −^1 n 2 +x 1 −^1 n j− 1 +x 1 −^1 n j+ 1 +···+x 1 −^1 n n n− 1 ≥ ( (x 1 x 2 ···xj− 1 xj+ 1 ···xn) n− ...

Algebra 379 This implies 1 8 − 1 4 + 1 2 (ab+bc+ca)−abc < 0 , and so 4(ab+bc+ca)− 8 abc≤1, and the desired inequality holds. ...

380 Algebra the expression. Repeating, we can decrease the expression to one in which all numbers are equal to^1 n. The value of ...

Algebra 381 E(a+α, b, c−α)=E(a, b, c)+α( 1 − 2 b)[(c−a)−α]. Sinceb≤canda+b+c=1, we haveb≤^12. This means thatE(a+α, b, c−α)≥ E(a ...

382 Algebra c=9,d=16, in which case the sum of the square roots is equal to 10. The inequality is proved. (V. Cârtoaje) 136.Ther ...

Algebra 383 while the product of the second pair is i^2 +im+ik+i(j− 1 )+(j− 1 )m+(j− 1 )k+(j− 1 )^2. We can see that the first o ...

384 Algebra Subtractingn(n 2 +^1 )from this expression, we obtain na^3 + 3 [ n(n+ 1 ) 2 −k ] a^2 + [ n(n+ 1 )( 2 n+ 1 ) 2 −^3 k ...

Algebra 385 Applying it fork=n=3,r 1 =r 2 =r 3 =3, and the numbersa 11 =a,a 12 =1,a 13 =1, a 21 =1,a 22 =b,a 23 =1,a 31 =1,a 32 ...

386 Algebra Multiplying allninequalities gives ∏n i= 1 ( n−xi 1 −xi ) ≤ ∏n i= 1 ( 1 + 1 n−√ (^1) x 1 ···xi− 1 xi+ 1 ···xn ) . Th ...

Algebra 387 Theorem.Ifa 1 ,a 2 ,a 3 ,b 1 ,b 2 ,b 3 are real numbers such that a 1 ≥a 2 ≥a 3 ≥ 0 ,b 1 ≥b 2 ≥b 3 ≥ 0 ,a 1 ≥b 1 ,a ...

388 Algebra xP 1 (x)=(x− 9 )P 1 (x+ 1 ). Arguing as before, we find thatP 1 (x)=(x− 1 )(x− 9 )P 2 (x). Repeating the argument, w ...

Algebra 389 of course is impossible. Hence only linear and constant polynomials satisfy the given relation. (Romanian Team Selec ...